What force is needed to turn the Earth Pole-over Pole. Eg. UK now on Equator?

Specifically, if someone could specify the Joules energy required to force the Earth to slowly rotate the poles end-to-end by say 1 degrees rotation, and by 5 degrees.
To be clear I mean to literally rotate the Earth (both poles in opposite directions), say Singapore will therefore be below the Equator.

I am writing three odd stories and this is a fact I would like to be sure of, instead of trying to work it out with formulas.

This is a more complicated scenario than you may have thought. If what you wanted to do was change the tilt of the Earths axis, that would be a fairly straight-forward calculation. You apply a torque to the Earth and you would induce a precession of the axis - the north pole would stay where it is relative to the Earth's surface, but the Earth would start to wobble, much like a top that spins and wobbles at the same time. But if you actually want to shift the axis relative to the Earth's surface - so that the North Pole shifts to a different location while maintaining a 24-hour day - that's a more complicated question. You would have to apply a counter-rotation to the Earth's spin and induce a second rotation at a different angle. The force required to do this depends on how quickly you want the change to occur. By rough calculation - a force of 10^18 Newtons applied at the earth's surface for 35 years could slow the earth's rotation to 0. Then apply that same amount to force at a different angle to get the Earth to spin about the new axis of rotation. That's a force equal to about 10^14 tons. If you're willing for this to take a few million years you get a more reasonable value for the force required.

I thought if a force is applied to a Gyro over a long time such as over 7 seconds, precession did not occur or was less. Based on some old US battleship usage. But I don't know much about this. I'm surprised about what you explained and will follow it up until I understand it. When the moon swings across the sky over an ocean, how much does the seawater actually move east to west? And also like to know the drift from say -50degree Latitude up toward the Equator if there's any. I know the moon mostly lifts the water but I think it must drag the water a lot during the passing toward the horizon. Is this right? Any idea of the rate or distance the water is typically dragged. Looks like an Oceanography question, but I just need a rough idea about this. I've looked online for many hours. Let me know about donation too. Very happy if you can help.

Specifically, if someone could specify the Joules energy required to force the Earth to slowly rotate the poles end-to-end by say 1 degrees rotation, and by 5 degrees.

The required force is based on how fast you wish to change the axis of the Earth's rotation and the angular velocity of the Earth. I've never worked with problems like this so I can't tell you how to do it right now and my head is immersed in something else right now.

When the moon swings across the sky over an ocean, how much does the seawater actually move east to west? And also like to know the drift from say -50degree Latitude up toward the Equator if there's any. I know the moon mostly lifts the water but I think it must drag the water a lot during the passing toward the horizon. Is this right? Any idea of the rate or distance the water is typically dragged.

I found a reference in this article: https://en.wikipedia.org/wiki/Tide that suggests the acceleration of water due to the lunar tide influence is maximum of 1.1 x 10^-7 g, which is approximately 1.1 x 10 ^-6 m/s. If we assume a sinusoidal function with 24 hour period, this works out to a max velocity reached of about 0.01 m/s, and amplitude of displacement of about 200 meters. This would be the average deep ocean movement. Obviously tides near shore typically move much quicker than that - they are greatly influenced by the topography of the shore.